Cho tam giác \(ABC\) cân tại \(A,\,\,\angle B = {65^0},\)  đường cao \(CH = 3,6\).  Hãy giải tam giác \(ABC\).

$\dfrac{{MN}}{{NP}}$

$\dfrac{{MP}}{{NP}}$

$\dfrac{{MN}}{{MP}}$

$\dfrac{{MP}}{{MN}}$

\(BH = 2\,cm\), \(CH = 3,2\,cm\), \(AC = 4\,cm\), \(AH = 2,4\,cm\)

\(BH = 1,8\,cm\), \(CH = 3,2\,cm\), \(AC = 4\,cm\), \(AH = 2,4\,cm\).

\(BH = 1,8\,cm\), \(CH = 3,2\,cm\), \(AC = 3\,cm\), \(AH = 2,4\,cm\)

\(BH = 1,8\,cm\), \(CH = 3,2\,cm\), \(AC = 4\,cm\), \(AH = 4,2\,cm\)

\(x = 14\)

\(x = 13\)

\(x = 12\)

\(x = \sqrt {145} \)

\({b^2} = b'.a\)

\(\dfrac{1}{{{h^2}}} = \dfrac{1}{{{c^2}}} + \dfrac{1}{{{b^2}}}\) 

\(a.h = b'.c'\)

\({h^2} = b'.c'\)

\(A{H^2} = BH.CH\)

\(A{B^2} = BH.BC\)

\(\dfrac{1}{{A{H^2}}} = \dfrac{1}{{A{B^2}}} + \dfrac{1}{{A{C^2}}}\)

\(AH.AB = BC.AC\)

\(a > b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}\)

\(a < b \Leftrightarrow \sqrt[3]{a} < \sqrt[3]{b}\)

\(a \ge b \Leftrightarrow \sqrt[3]{a} \ge \sqrt[3]{b}\)

\(a < b \Leftrightarrow \sqrt[3]{a} > \sqrt[3]{b}\)

\(\sqrt {2018 + 2019}  = \sqrt {2018}  + \sqrt {2019} \)

\(\sqrt {2018. 2019}  = \dfrac{{\sqrt {2018} }}{{\sqrt {2019} }}\)

\(\sqrt {2018} .\sqrt {2019}  = \sqrt {2018.2019} \)

\(2018. 2019 = \dfrac{{\sqrt {2019} }}{{\sqrt {2018} }}\)